Concept explainers
The mass of lead should be added to give the diver neutral buoyancy.
Answer to Problem 49P
3.9 kg
Explanation of Solution
Given:
Mass of a diver, m=85 kg
Density of a diver, ρ=0.96 kg/L
Formula used:
Let, volume of the diver be V
Density of the diver be ρD
Volume of the added lead be VPb
Mass of the lead be mPb
Let us apply ∑Fy=0 to the diver,
B−WD−WPb=0 (1)
Where, B is buoyant force
B=ρwVD+PbgB=ρwVDg+ρwVPbg
WD is weight of a diver
WD=mDg=ρDVDg
WPb is weight of the lead, WPb=mPbg
Substituting for B , WD , and WPb in equation (1) ,
ρwVDg+ρwVPbg−ρDVDg−mPbg=0
Or
ρwVD+ρwVPb−ρDVD−mPb=0 (2)
Where, ρw is density of water,
VD is volume of a diver,
g is acceleration due to gravity,
VPb is volume of the lead,
ρD is density of a diver,
mPb is mass of a lead.
Mass (m) , density (ρ) , and volume (V) of a given substance is related through the expression,
m=ρV⇒ρ=mV
Applying this to the equation (2) ,
ρwmDρD+ρwmPbρPb−ρDmDρD−mPb=0
⇒ρwmDρD−mD+ρwmPbρPb−mPb=0
⇒mD(ρwρD−1)+mPb(ρwρPb−1)=0
⇒mD(ρw−ρDρD)+mPb(ρw−ρPbρPb)=0
⇒mPb(ρw−ρPbρPb)=−mD(ρw−ρDρD)
or
−mPb(ρPb−ρwρPb)=−mD(ρw−ρDρD)
or
mPb(ρPb−ρwρPb)=mD(ρw−ρDρD)
⇒mPb=mD(ρw−ρDρD)(ρPb−ρwρPb)=ρPb(ρw−ρD)mDρD(ρPb−ρw) (3)
Calculation:
Density of lead, ρPb=11.3×103kg/m3
Density of water, ρw=1.0×103kg/m3
Substituting the numerical values in equation (3) ,
mPb=(11.3×103kg/m3)(1×103kg/m3-0.96×103kg/m3)85 kg(0.96×103kg/m3)(11.3×103kg/m3-1×103kg/m3)=3.9 kg
Conclusion:
The mass of lead should be added to give the diver neutral buoyancy 3.9 kg .
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Chapter 13 Solutions
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